For two probability measure $\mu$ and $\nu$ on $\mathbb{R}$, we call $\mu$ is *stochastically smaller* than $\nu$ (i.e., $\mu\leq\nu$) , if $\int f \, d\mu\leq\int f \, d\nu$ for any nonnegative bounded increasing measurable function $f:\mathbb{R}\to\mathbb{R}$.

There is a fact that $\mu\leq\nu$ is equivalent to $F_\mu(x)\geq F_\nu(x)$ for all $x\in\mathbb{R}$, where $F_\mu(x)$ and $F_\nu(x)$ are the cumulative distribution functions of $\mu$ and $\nu$ respectively.

Another fact is that, $\mu\leq\nu$ is equivalent to $\int f \, d\mu\leq\int f \, d\nu$ for any nonnegative bounded increasing measurable function $f:\mathbb{R}\to\mathbb{R}$ with Lipschitz constant less or equal to $1.$

I'm curious about whether there is a *strong order* in this ordered space. That is to say, can we find some topology in probability space such that, there exists two probability measure $\mu$ and $\nu$, and their open neighborhood $\mu\in U$, $\nu\in V$ in this topology such that, $\tilde{\mu}\leq \tilde{\nu}$, for any $\tilde{\mu}\in U$ and $\tilde{\nu}\in V$.

I guess we cannot find such open sets in weak convergence topology, or Wasserstein metric $W_p$, or total variation distance topology. I cannot strictly prove it. The intuition tells me that for any $\mu\leq\nu$, perturbate the distribution function of $\mu$ and $\nu$ such that it changes little near $-\infty$ will cause contradiction.

The same question also exists for other partial order in the probability space.

Any advices will be greatly appreciated.